# Write the general term in expansion of (x² - yx)¹², x ≠ 0

**Solution:**

It is known that (r + 1)^{th} term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)^{n} is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ

Thus, the general term in the expansion of (x² - yx)¹² is

Tᵣ ₊ ₁ = ¹²Cᵣ(x²)¹² ⁻ ʳ (- yx)ʳ

= ¹²Cᵣ(x)²⁴ ⁻ ²ʳ (- 1)ʳ (y)ʳ (x)ʳ

= (- 1)ʳ ¹²Cᵣ(x)²⁴ ⁻ ʳ (y)ʳ

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 4

## Write the general term in expansion of (x² - yx)¹², x ≠ 0

**Summary:**

Using the binomial theorem, the general term in the expansion of (x² - yx)¹², x ≠ 0 is to be found. We have found that it is equal to (- 1)ʳ ¹²Cᵣ(x)²⁴ ⁻ ʳ (y)ʳ